# Initial and Final Values Properties of The Laplace Transform

The initial-value and final-value properties allow us to find the initial value $f(0)$ and the final value $f(\infty)$ of $f(t)$ directly from its Laplace transform $F(s)$. To obtain these properties, we begin with the differentiation property in Eq. (1), namely,
$$L[f′(t)]=sF(s)−f(0_−) \tag{1}$$
$$s F(s)-f\left(0^{+}\right)=\mathcal{L}\left[\frac{d f}{d t}\right]=\int_{0}^{\infty} \frac{d f}{d t} e^{-s t} d t \tag{2}$$
If we let $s \rightarrow \infty$, the integrand in Eq. (2) vanishes due to the damping exponential factor, and Eq. (2) becomes
$$\lim _{s \rightarrow \infty}\left[s F(s)-f\left(0^{+}\right)\right]=0$$
or
$$f\left(0^{+}\right)=\lim _{s \rightarrow \infty} s F(s)$$
This is known as the initial-value theorem. For example,
$$f(t)=e^{-2 t} \cos 10 t \quad \Longleftrightarrow \quad F(s)=\frac{s+2}{(s+2)^{2}+10^{2}}$$
Using the initial-value theorem,
\begin{aligned}f\left(0^{+}\right)=\lim _{s \rightarrow \infty} s F(s) &=\lim _{s \rightarrow \infty} \frac{s^{2}+2 s}{s^{2}+4 s+104} \\&=\lim _{s \rightarrow \infty} \frac{1+2 / s}{1+4 / s+104 / s^{2}}=1\end{aligned}
which confirms what we would expect from the given $f(t)$. In Eq. (2), we let $s \rightarrow 0$; then
$$\lim _{s \rightarrow 0}\left[s F(s)-f\left(0^{-}\right)\right]=\int_{0}^{\infty} \frac{d f}{d t} e^{0 t} d t=\int_{0}^{\infty} d f=f(\infty)-f\left(0^{-}\right)$$
or
$$f(\infty)=\lim _{s \rightarrow 0} s F(s) \tag{3}$$
Fig. 1
This is referred to as the final-value theorem. In order for the final-value theorem to hold, all poles of $F(s)$ must be located in the left half of the $s$ plane (see Fig. 1); that is, the poles must have negative real parts. The only exception to this requirement is the case in which $F(s)$ has a simple pole at $s=0$, because the effect of $1 / s$ will be nullified by $s F(s)$ in Eq. (3). For example,
$$f(t)=e^{-2 t} \sin 5 t \quad \Longleftrightarrow \quad F(s)=\frac{5}{(s+2)^{2}+5^{2}}$$
Applying the final-value theorem,
$$f(\infty)=\lim _{s \rightarrow 0} s F(s)=\lim _{s \rightarrow 0} \frac{5 s}{s^{2}+4 s+29}=0$$
as expected from the given $f(t)$. As another example,
$$f(t)=\sin t \quad \Longleftrightarrow \quad f(s)=\frac{1}{s^{2}+1}$$
so that
$$f(\infty)=\lim _{s \rightarrow 0} s F(s)=\lim _{s \rightarrow 0} \frac{s}{s^{2}+1}=0$$
This is incorrect, because $f(t)=\sin t$ oscillates between $+1$ and $-1$ and does not have a limit as $t \rightarrow \infty$. Thus, the final-value theorem cannot be used to find the final value of $f(t)=\sin t$, because $F(s)$ has poles at $s=\pm j$, which are not in the left half of the $s$ plane.
In general, the final-value theorem does not apply in finding the final values of sinusoidal functions-these functions oscillate forever and do not have final values. The initial-value and final-value theorems depict the relationship between the origin and infinity in the time domain and the $s$ domain. They serve as useful checks on Laplace transforms.